ĐKXĐ: \(x\ge0\)
\(B=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(B_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)
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